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Difficulty: Very hardPolygon Angles and Properties

A convex polygon has nn sides. The interior angles of the polygon form an arithmetic progression with a common difference of d∘d^\circ, where dd is a positive integer. If the smallest interior angle of the polygon measures 100∘100^\circ, what is the maximum possible value of nn?

Answer: 8

Answer

The maximum possible value of nn is 8.
The correct answer is 8 because we set the sum of the interior angles (n−2)×180∘(n-2) \times 180^\circ equal to the sum of the arithmetic progression n2[200+(n−1)d]\frac{n}{2}[200 + (n-1)d]. Solving for (n−1)d(n-1)d gives (n−1)d=160−720n(n-1)d = 160 - \frac{720}{n}. Since the polygon is convex, the largest angle 100+(n−1)d100 + (n-1)d must be strictly less than 180∘180^\circ, which means (n−1)d<80∘(n-1)d < 80^\circ. Substituting the expression yields 160−720n<80160 - \frac{720}{n} < 80, which simplifies to n<9n < 9. Since nn must be an integer, the maximum possible value of nn is 8. For n=8n = 8, the common difference d=10d = 10 is a positive integer, satisfying all conditions.

Step-by-Step Solution

1
Express the sum of the interior angles using the polygon angle sum formula and the arithmetic progression formula.
The sum of the interior angles of a convex polygon with nn sides is (n−2)×180∘(n-2) \times 180^\circ. The sum of the angles in arithmetic progression with first term 100∘100^\circ and common difference d∘d^\circ is n2[2(100)+(n−1)d]=100n+n(n−1)d2\frac{n}{2}[2(100) + (n-1)d] = 100n + \frac{n(n-1)d}{2}.
This sets up the system relating the geometry of the polygon to the algebraic progression of its angles.
2
Equate the two expressions and solve for the quantity (n−1)d(n-1)d.
100n+n(n−1)d2=180n−360  ⟹  n(n−1)d2=80n−360  ⟹  (n−1)d=160−720n100n + \frac{n(n-1)d}{2} = 180n - 360 \implies \frac{n(n-1)d}{2} = 80n - 360 \implies (n-1)d = 160 - \frac{720}{n}.
This isolates the quantity (n−1)d(n-1)d, which represents the difference between the largest and smallest angles.
3
Apply the convexity constraint that every interior angle must be strictly less than 180∘180^\circ.
The largest angle is the last term of the progression: 100+(n−1)d100 + (n-1)d. For the polygon to be convex, this angle must be strictly less than 180∘180^\circ. Therefore, 100+(n−1)d<180  ⟹  (n−1)d<80100 + (n-1)d < 180 \implies (n-1)d < 80. Substituting (n−1)d=160−720n(n-1)d = 160 - \frac{720}{n} gives 160−720n<80  ⟹  80<720n  ⟹  n<9160 - \frac{720}{n} < 80 \implies 80 < \frac{720}{n} \implies n < 9.
A convex polygon cannot have any interior angles greater than or equal to 180∘180^\circ.
4
Identify the maximum integer value of nn and verify that a positive integer common difference dd exists.
Since n<9n < 9 and nn must be an integer, the maximum possible value is n=8n = 8. For n=8n = 8, we calculate (8−1)d=160−7208  ⟹  7d=70  ⟹  d=10(8-1)d = 160 - \frac{720}{8} \implies 7d = 70 \implies d = 10. Since d=10d = 10 is a positive integer, the solution is valid.
This ensures the result satisfies all constraints, including that the common difference is a positive integer.

Key Concept

The sum of the interior angles of a convex polygon with nn sides is (n−2)×180∘(n-2) \times 180^\circ, and all interior angles of a convex polygon must be strictly less than 180∘180^\circ.
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Topics

Polygon Angles and Properties

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